= Solution
Let $M$ denote an incident <muon> and $S$ a detector signal. The selected sample is conditioned on $S$. The <law of total probability> gives
$$
P(S)=P(S\mid M)P(M)+P(S\mid M^c)P(M^c)
=(0.90)(0.95)+(0.20)(0.05)=0.865.
$$
By <Bayes' theorem>, the contamination, namely the <conditional probability> that a selected particle is not a <muon>, is
$$
\boxed{P(M^c\mid S)=\frac{(0.20)(0.05)}{0.865}
=0.01156\simeq1.16\%.}
$$
For example, per $10\,000$ incident particles the expected selected counts are $8550$ true <muons> and $100$ other particles. The denominator is the $8650$ selected particles; the rejected <muons> are outside this sample.
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